Unit 9 – Chemical Quantities

1. Unit 9 – Chemical Quantities Chemical EquationsMole-Mole RelationshipsMass CalculationsLimiting ReactantPercent Yieldpages 246-266

2. Chemical Quantities Upon completion of this unit, you should be able to do the following: • Determine relationships between moles of reactants and moles of products from a balanced equation. • Calculate the mass, moles, gas or solution volume of a reactant or product, given the mass, moles, gas or solution volume of another reactant or product. • Use the limiting reactant to do stoichiometric calculations. • Calculate actual yield as a percentage of theoretical yield.

3. Chemical Equations Pg 247-249 Suppose that after you graduate from college, you go to work for a chemical company. You are working with a chemist to improve the process that produces methanol from the reaction of hydrogen and carbon monoxide. Your first assignment is to order enough hydrogen and carbon monoxide to produce 6 kg of methanol. How would you determine how much of each material to order?

4. Chemical Equations Write the unbalanced equation for the reaction. CO (g) + H2 (g) → CH3OH (l) Next, balance the equation. CO (g) + 2 H2 (g) → CH3OH (l) Understand that you can multiply this equation by any number and it is still balanced. For example, multiply by 12 to get 12 CO (g) + 24 H2 (g) → 12 CH3OH (l) This is still balanced. Since 12 is a dozen we could also write this as 1 dozen CO (g) + 2 dozen H2 (g) → 1 dozen CH3OH (l)

5. Chemical Equations We could also multiply the equation by a very large number, say 6.022 x 1023. 6.022 x 1023 [CO (g) + 2 H2 (g) → CH3OH (l)] or 6.022 x 1023 CO (g) + 2 (6.022 x 1023 ) H2 (g) → 6.022 x 1023 CH3OH (l) Just as 12 is called a dozen, 6.022 x 1023 is called a mole so we could rewrite the equation as 1 mol CO (g) + 2 mol H2 (g) → 1 mol CH3OH (l)

6. Chemical Equations Example 9.1, page 248 Propane is a fuel commonly used for cooking on gas grills. Propane reacts with oxygen to produce heat and the products of carbon dioxide and water vapor. This combustion reaction is represented by the unbalanced equation C3H8 (g) + O2 (g) → CO2 (g) + H2O(g) Give the balanced equation and state the meaning of the equation in terms of moles. 1C3H8 (g) + 5O2(g) → 3CO2 (g) + 4H2O (g) 1 mol of C3H8 reacts with 5 mol of O2 to form 3 mol of CO2 and 4 mol of H2O

7. Mole-Mole Relationships Now use the balanced equation to predict the moles of products that a given number of reactants will yield. Consider the decomposition of water into hydrogen and oxygen. 2 H2O (g) → 2 H2 (g) + O2 (g) The equation tells you that 2 mol of H2O will decompose to yield 2 mol of H2 and 1 mol of O2.

8. Mole-Mole Relationships 2 H2O (g) → 2 H2 (g) + O2 (g) How much H2 and O2 would be produced from 4 mol of H2O? How much H2 and O2 would be produced from 5.8 mol of H2O? 2 H2O (g) → 2 H2 (g) + O2 (g) To answer, first divide all coefficients by 2 1 H2O (g) → 1 H2 (g) + ½ O2 (g) (non-integer coefficients only make sense for moles)

9. Mole-Mole Relationships How much H2 and O2 would be produced from 5.8 mol of H2O? 2 H2O (g) → 2 H2 (g) + O2 (g) To answer, first divide all coefficients by 2 1 H2O (g) → 1 H2 (g) + ½ O2 (g) (non-integer coefficients only make sense for moles) Next, multiply all coefficients by 5.8 5.8 H2O (g) → 5.8 H2 (g) + 2.9 O2 (g)

10. Mole Ratios An easier way to solve this is to use mole ratios. How many mol of O2 would be produced from 5.8 mol of H2O? 2 H2O (g) → 2 H2 (g) + O2 (g) From the equation for the reaction we can see that 2 mol H2O yield 1 mol O2 . Written as a ratio it is 1 mol O2 2 mol H2O So 5.8 mol H2O would yield 5.8 mol H2O x 1 mol O2 = 2.9 mol O2 2 mol H2O

11. Mole Ratios Page 251, Example 9.3 Calculate the number of moles of oxygen required to react exactly with 4.3 mol of propane per the balanced equation C3H8 (g) + 5 O2(g) → 3 CO2(g) + 4 H2O (g) Answer: 21.5 mol O2(g) Homework: page 269, problem 9.16

12. Mass Calculations • Moles represent the number of molecules and we cannot measure molecules directly. In chemistry we count by weighing, so we need to convert from moles to mass in our calculations. • Consider the synthesis reaction between aluminum and iodine. 2 Al (s) + 3I2 (s) → 2 AlI3 (s) • Suppose we have 35.0 g of aluminum. How much iodine should we weigh out to react exactly with this amount of aluminum?

13. Mass Calculations 2 Al (s) + 3I2 (s) → 2 AlI3 (s) 3 moles I2 2 moles Al moles Al present x 3 moles I2 = moles I2 needed 2 moles Al 1 mol Al x 3 moles I2 = 1.95 moles I2 x 253.8 gI2 26.98 g Al 2 moles Al 1 mol I2 = 495 g I2

14. Mass Calculations • Propane, C3H8 , when used as a fuel, reacts with oxygen to produce carbon dioxide and water according to the following unbalanced reaction C3H8 (g) + O2 (g) → CO2 (g) + H2O(g) • What mass of oxygen will be required to react exactly with 96.1 g of propane? Answer: 349 g O2 Homework: page 270, problem 26

15. Limiting Reactant • Lab 16 – Limiting Reactant (Reagent) • Reactant is a substance that is consumed in a reaction • Reagent is a substance that is added to a system in order to bring about a chemical reaction • A solvent or catalyst would be a reagent without being a reactant. • In Lab 16, potassium iodate and barium chloride are both reactants and reagents. • Homework: page 271, problem 46

16. Percent Yield • The amount of product calculated based on a limiting reactant is called the theoretical yield. • It is the maximum amount of product that can be formed. • The actual yield is the amount of product actually formed. • The percent yield is the ratio of actual yield to theoretical yield times 100. actual yield x 100 = percent yield theoretical yield

17. Percent Yield Methanol can be produced by the reaction between carbon monoxide and hydrogen. 2 H2 (g) + CO (g) → CH3OH (l) Suppose 68.5 kg of CO (g) is reacted with 8.6 kg of H2 (g). • Calculate the theoretical yield of CH3OH (l). 68.6 kg CH3OH • If 35.7 kg of CH3OH (l) is actually produced, what id the percent yield? 52.0% yield Homework: page 266, self check 9.7